A sample of 81 observations has a mean 23 and standard deviation
3.6, Test the Null Hypothesis H0 : μ = 25 against the alternative
H1: μ < 25, assuming α = .01
"n=81 \\\\\n\n\\bar{x} = 23 \\\\\n\ns = 3.6 \\\\\n\nH_0: \\mu = 25 \\\\\n\nH_1: \\mu < 25"
Test-statistic
"t= \\frac{\\bar{x} - \\mu}{s \/ \\sqrt{n}} \\\\\n\nt = \\frac{23-25}{3.6 \/ \\sqrt{81} } \\\\\n\n= \\frac{-2}{0.4333} \\\\\n\n= -1.566 \\\\\n\n\u03b1=0.01 \\\\\n\ndf=n-1=81-1=80"
One-tailed test
"t_{crit} = 2.373"
Reject the null hypothesis if "t \u2264 -t_{crit}"
"t= -1.566 > t_{crit} = -2.373"
We accept the null hypothesis at 0.01 level of significance.
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